Given a binary search tree (BST), find the lowest common ancestor (LCA) of two given nodes in the BST.

According to the : “The lowest common ancestor is defined between two nodes v and w as the lowest node in T that has both v and w as descendants (where we allow a node to be a descendant of itself).”

_______6______       /              \    ___2__          ___8__   /      \        /      \   0      _4       7       9         /  \         3   5 

For example, the lowest common ancestor (LCA) of nodes

 

2

 

and

 

8

 

is

 

6

. Another example is LCA of nodes

 

2

 

and

 

4

 

is

 

2

, since a node can be a descendant of itself according to the LCA definition.

来源： <>

  

思路，先判断入口是否有非法输入。

 

1 如果某一个root==p || root == q，那么LCA肯定是root（因为是top down，LCA肯定在root所囊括的树上，而root又是p q其中一个节点了，那么另外一个节点肯定在root之下，那么root就是LCA），那么返回root

 

2 如果root<min(p, q)，那么LCA肯定在右子树上，那么递归

 

3 如果max(p, q)<root，那么LCA肯定在左子树上，那么递归

 

4 如果p<root<q，那么root肯定为LCA

1. /**
2. * Definition for a binary tree node.
3. * struct TreeNode {
4. * int val;
5. * TreeNode *left;
6. * TreeNode *right;
7. * TreeNode(int x) : val(x), left(NULL), right(NULL) {}
8. * };
9. */
10. class Solution {
11. public:
12. TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
13. if(root==p||root==q)
14. return root;
15. if(p->val>q->val)
16. swap(p,q);
17. if(root->val > p->val && root->val < q->val)
18. return root;
19. TreeNode *node;
20. if(root->val > max(p->val,q->val))
21. node=lowestCommonAncestor(root->left,p,q);
22. if(root->val < min(p->val,q->val))
23. node=lowestCommonAncestor(root->right,p,q);
25. return node;
26. }
28. };